Question

3. The outdoor air enters an AC system at 10°C and 75% relative humidity at a steady rate of 26 m3/ min, and it leaves at 25°C and 45% relative humidity. The outdoor air is first heated to 21°C in the heating section and then humidified by injection of hot steam in the humidifying section. Assuming the entire process takes place at 1 atm, determine (a) the rate of heat supply in the heating section, and (b) the mass flow rate of steam required in the humidifying section. (Answers: 5.89 kW; 0.0017 kg/s)

Answer 1

Given data 1

T = 10 °C

01 = 0.75

T3 25 °C

03 = 0.45

T2 = 21 °C

Diagram →

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Solution →

Applying the mass and energy balances on the heating section gives

mass balance →

mai ma2 = ma

water mass balance

malwi = ma2w2

$\omega _{1}=\omega _{2}$

Energy balance

Q=m* (h2 - hu)

saturation pressure at 10 °C

Paat 1.2282 kPa

1g1 = 2520.42 kJ/kg

Relative humidity is given by

Pu 01 sat

0.75 Pu 1.2282

Py = 0.92115 kPa

01 RT1 P-P.

11 0.287 * (10 + 273 101.325 - 0.92115

V1 0.8089 m/kg

V ma U1

26 ma 60 * 0.8089

ma 0.5356 kg/s

اله 0.622P P - P 11

الناس 0.622 0.02115 101.325 - 0.902115

الد 5.7065 * 103 kg/kg of dry air

kg / kg of dry air 108 * 5.7065 = ال = ال =

h1 = CpTi + wihg1

h1 = 1.005 * 10+ 5.7065 * 10-3* 2520.42

hi 24.43 kJ/kg

hg2 = 2540.65 kJ/kg at 21 °C

h2 = CpT2 +wohg2

h2 = 1.005 * 21 +5.7065 * 10-3* 2540.65

h2 = 35.60 kJ/kg

Energy balance

Q=m* (h2 - hu)

Q = 0.5356 * 35.60 – 24.43)

Q = 5.98 kW

saturation pressure at 25 °C

Psat = 3.1717 kPa

Relative humidity is given by

Pu 03 sat

0.45 Py 3.1717

Py = 1.4272 kPa

$\omega _{3}=\frac{0.622P_{v}}{P-P_{v}}$

$\omega _{3}=\frac{0.622*1.4272}{101.325-1.4272}$

$\dot{m}_{w}=\dot{m}_{a}*\left ( \omega _{3}-\omega _{1} \right )$

$\dot{m}_{w}=1.7033*10^{-3}\textup{ kg/s}$